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mathmath333

  • one year ago

functions

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\normalsize \text{Find the maximim value of the function}\hspace{.33em}\\~\\ & \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)

  2. SolomonZelman
    • one year ago
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    interval given ?

  3. SolomonZelman
    • one year ago
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    with out interval this maximum value diverges to infinity.

  4. mathmath333
    • one year ago
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    this question means that i have to minimize \(\large \color{black}{\begin{align} x^2-3x+2 \hspace{.33em}\\~\\ \end{align}}\) right ?

  5. SolomonZelman
    • one year ago
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    should be, or els... i mean there is not maximum value here, if you are taking over \((-\infty,~+\infty)\)

  6. SolomonZelman
    • one year ago
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    wait

  7. SolomonZelman
    • one year ago
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    1/that ? or, just that ?

  8. mathmath333
    • one year ago
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    lol its \(\large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)

  9. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\) and you want to find absolute minimum, right?

  10. SolomonZelman
    • one year ago
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    well, I can tell you that the limit as x approaches \(\pm\)infinity is going to be 0.

  11. SolomonZelman
    • one year ago
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    we can plot, I am guessing it is something like |dw:1435010239497:dw|

  12. mathmath333
    • one year ago
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    i want to find the maximum value of this \(\large \color{black}{\begin{align} \dfrac{1}{x^2-3x+2} \hspace{.33em}\\~\\ \end{align}}\)

  13. SolomonZelman
    • one year ago
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    none

  14. SolomonZelman
    • one year ago
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    it is +infinity

  15. mathmath333
    • one year ago
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    but how ?

  16. SolomonZelman
    • one year ago
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    https://www.desmos.com/calculator

  17. mathmath333
    • one year ago
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    how i prove that with algebra

  18. SolomonZelman
    • one year ago
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    even the minimum doesn't exist.

  19. SolomonZelman
    • one year ago
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    with algebra, even with no calc ?

  20. mathmath333
    • one year ago
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    idk calculus

  21. SolomonZelman
    • one year ago
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    ic

  22. SolomonZelman
    • one year ago
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    lets c

  23. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-3x+2} }\) it has vertical asymptotes, can you find them for me?

  24. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} x=2,\ 1 \hspace{.33em}\\~\\ \end{align}}\)

  25. SolomonZelman
    • one year ago
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    yes

  26. SolomonZelman
    • one year ago
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    and, the best thing I can see here is to use the idea of a limit to see where range tends to the close we get to x=1 and x=2. I don't really know a purely algebraic proof for why this function has no max or min.

  27. xapproachesinfinity
    • one year ago
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    that has a max actually

  28. SolomonZelman
    • one year ago
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    yes?

  29. SolomonZelman
    • one year ago
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    https://www.desmos.com/calculator keep scrollin up

  30. SolomonZelman
    • one year ago
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    on a certain interval, no doubt (unless it has a domain gap)

  31. xapproachesinfinity
    • one year ago
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    i think it attains one max does not perhaps required any domain to get it or i think

  32. xapproachesinfinity
    • one year ago
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    your graph does not show up see here https://www.desmos.com/calculator

  33. SolomonZelman
    • one year ago
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    i don't think there is a max. that is due to an asymptote - two of them, there.

  34. SolomonZelman
    • one year ago
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    we can find the limit from the right and from the left as x approaches 1 and 2.

  35. xapproachesinfinity
    • one year ago
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    http://prntscr.com/7k4q1l here see that bottom part of the graph

  36. xapproachesinfinity
    • one year ago
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    with calculus we can prove it has a max value but @mathmath333 is looking for a different way

  37. SolomonZelman
    • one year ago
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    this graph goes further down and up if you scroll on desmos, but we can go ahead and take any limit that is closed from either side of any of the asymptotes you wish to choose.....

  38. xapproachesinfinity
    • one year ago
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    i mean here local max not abolute of course

  39. mathmath333
    • one year ago
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    how to find that local max

  40. SolomonZelman
    • one year ago
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    oh, but the question is asking for absolute doesn't it?

  41. SolomonZelman
    • one year ago
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    that is what i would think of course

  42. xapproachesinfinity
    • one year ago
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    first let's write that as 1/(x-1)(x-2)

  43. xapproachesinfinity
    • one year ago
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    i had thought of absolute too, just not sure about it!

  44. mathmath333
    • one year ago
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    yes the question is asking absolute, i just asked that of curiosity

  45. SolomonZelman
    • one year ago
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    this function is pretty weird. I wonder what would it model

  46. SolomonZelman
    • one year ago
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    i mean in real world

  47. xapproachesinfinity
    • one year ago
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    you want to then prove there is no max algebraically

  48. mathmath333
    • one year ago
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    yes ^

  49. SolomonZelman
    • one year ago
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    would that include using ideas from calc but not its techniques?

  50. SolomonZelman
    • one year ago
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    I am stubborn on the idea of a limit. A limit would simply show that as it approaches __ it can be infinitely large/small

  51. mathmath333
    • one year ago
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    ok i will try to bear with calculus

  52. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\) we need to do only 1 asymptote for this

  53. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{1 }{x^2-3x+2} }\) lets plug in values that are less than, but close to 1

  54. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f(0.9)=? }\)

  55. SolomonZelman
    • one year ago
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    = about 9

  56. SolomonZelman
    • one year ago
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    f(0.99)=99

  57. SolomonZelman
    • one year ago
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    try plugging it yourself for close values to 1

  58. SolomonZelman
    • one year ago
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    these values are called approaching "from the left" because they are a buit less than 1

  59. SolomonZelman
    • one year ago
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    now, try to do f(1.5) f(1.1) f(1.01) f(1.001) and you will see where it goes "from the right" - for values that are greater than 1

  60. SolomonZelman
    • one year ago
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    i got to go

  61. SolomonZelman
    • one year ago
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    sorry

  62. xapproachesinfinity
    • one year ago
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    suppose there exist a max max(1/(x^2-3x+2)=c 1/x^2-3x+2<c x^2-3x+2>1/c=l x^2-3x+2-l>0

  63. xapproachesinfinity
    • one year ago
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    just some random stuff lol

  64. xapproachesinfinity
    • one year ago
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    coming back..

  65. mathmath333
    • one year ago
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    ok i need to go please comment if u found something

  66. xapproachesinfinity
    • one year ago
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    let's first consider the domain \(D_f: \{ x\in \mathbb{R}: x\ne1, x\ne 2 \}\)

  67. xapproachesinfinity
    • one year ago
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    that is (-oo, 1)u(1, 2)u(2,oo)

  68. xapproachesinfinity
    • one year ago
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    we look at the sign of y when x<1 if y does not change the sign in that interval then i must not have a max seems that 1/(x-2)(x-1)>0 for x<1 same reasoning for x>2 ===> y>0 so only btw 1 and 2 left

  69. xapproachesinfinity
    • one year ago
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    from the graph we knew for 1<x<2, y<-4 we need to show this

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